微积分第三章习题参考答案

6
(t2
t
1
t
1
)dt 1
2t 3 3t 2 6t 6 ln(t 1) c
2 x 1 33 x 1 66 x 1
ln( 6 x 1 1) c.
p54.4.解 法 1:
1
x4 1 x4
I
x3(x4
dx 1)
x3( x4 1) dx
1 ( x3
x x 4 1 )dx
(e xch x ) e x (ch x sh x ) e 2 x ,
( 1 e 2 x ) (e x sh x ) (e xch x ) e 2 x , 2
1 e 2 x , e x sh x , e xc h x 都 是 e 2 x的 原 函 数 . 2
§3.2不定积分的换元法(53-54)
2
4
3. xf ( x) f ( x) c . 4. x ln x x c .
5. x arcsin x 1 x2 c .
6. 1 x cos 2 x 1 sin 2 x c 1 x sin2 x x sin 2 x c .
4
8
2
48
x
sin
x
cos
xdx
1 2
x
sin
2 xdx
x l n x d x 1 l n x d x 1
1
x 1 l n x 1
xdx
1 1
x 1 ln x
x 1
1 ( 1)2 c .
p56. 7.I0 xc,
In xlnn xn lnn1 xdxxlnn xnIn1
§3.4 有理函数的不定积分(57-58)
1 x2
p60.
6.
cos4
xdx
(1
cos 2 x 2
)2dx
1 4
(1
2cos 2 x
cos2
2 x)dx
1 4
[1
2 cos
2
x
wenku.baidu.com
1 2
(1
cos 4
x )]dx
3 x cos x 3 x ln 3d sin x
3 x cos x 3 x sin x ln 3 3 x sin x ln 2 3dx
3x cos x 3x sin x ln 3 I ln2 3,
I
1
1 ln 2
3
(3x
sin
x
ln
3
3x
cos
x)
c.
p 5 6 . 4 .令 u x ,则 x u 2 , d x 2 u d u .
3.
x2
2x
5
(x
1)2
4
arctan 2
2
c.
p57.二 .1.
(4 x 6)dx x3 5x2 6x
(
1 x
6 x
3
x
7
)dx 2
ln x 6 ln( x 3) 7 ln( x 2) c.
2.
x 3dx x3
(x2
3x
9
27 )dx x3
x3 3x2 9 x 27 ln | x 3 | c.
sin(ln
x )dx
1 [ x sin(ln 2
x)
x cos(ln
x)]
c.
p 6 0 . 5 .
e arctan x
u arctan x
dx
(1 x 2 )3
eu cos udu
1 e u (sin u cos u) c 2
1 e arctan x 1 x c .
2
( x e x e x c1 ) |x0 ( x e x e x c 2 ) |x0 ,
c1 c2,
x e |x |d x
c
ex(x
e
x(
x
1)
1)
x0 .
x0
p 5 6 . 6 . 当 1时 ,
x l n x d x x 1 l n x d x 1 l n 2 x c ; 2 当 1时 ,
2.
x 2 arctan
xdx
1 3
arctan
xdx 3
1 x 3 a r c t a n x 1
3
3
x
3
1
1
x
2
d
x
1 x 3 arctan x 1 x 2 1 ln (1 x 2 ) c .
3
66
p56. 3.I 3x sin xdx 3x d cos x
3 x cos x cos x 3 x ln 3dx
2 s in x d e sin x 2 e sin x s in x 2 e sin x d s in x
2 e sin x sin x 2 e sin x c .
2.
m
ax{ x 2 ,1}dx
x2
d
d x
x
x 2 d x
x 1 | x | 1 x1
(接 上 页 )
8. cot x tan x c 2 csc 2 x c
2cot2 2 x c .
p52. shx e x ex ,chx e x ex .
2
2
四 .证 明 :
( 1 e 2 x ) e 2 x , 2
(e x sh x ) e x ( sh x ch x ) e 2 x ,
微积分第三章习题参考答案
§3.1不定积分的概念及计算(51-52)
p 5 1 .一 .1 .
f ( x )dx .
x
2. e 3 .
3. c .
4. ln | x | arctan x c .
5. csc x cot x c tan x c . 2
6. 1 ( x sin x ) c . 7. e x 2 x c . 2
2
3
4
p 5 3 .3 . I c o s 4 x (1 c o s 2 x ) 2 d c o s x
( c o s 4 x 2 c o s 6 x c o s 8 x ) d c o s x
2 cos7 x 1 cos5 x 1 cos9 x c.
7
5
9
4 . I ( s e c 2 x 1 ) 2 s e c 2 x d s e c x
2
2
p54.三 .1. 令 x a sin t ,
x 2dx a 2 sin2 tdt a2 x2
a2
a2 1
2
(1 cos 2t )dt
2
(t sin 2t) c 2
a2
x x a2 x2
arcsin
c.
2
a
2
p 5 4 .2 .解 法 ( 1 ) d x
x x2 4
1 3
x
3
c1
x 1
x c | x | 1 ,
1 3
x3 c2
x1
由
1 3
x3
c1
x 1 0
x
c
x10 ,得 c1
c
2, 3
由
1 3
x3
c1
x 1 0
x
c
x10 ,得 c2
c
2, 3
(接上页)
1 3
x3
2 3
c
max{| x |,1}dx x c
x 1 | x | 1.
p 5 3 .6 . I
da tan t
(a 2 tan 2 t a 2 )3
a sec2 tdt a 3 sec3 t
1 a2
cos tdt
1 a2
sin
t
c
a2
x
c.
a2 x2
二 .1. 2.
I 1 e 2x2 c; 4
I
dx 9 4x2
xdx 9 4x2
1 arcsin 2 x 9 4 x 2 c .
1 4
xd
cos 2 x
1 4
x
cos
2
x
1 4
cos
2
xdx
1 4
x
cos
2
x
1 8
sin
2
x
c.
p 5 5 .二 1 . x 2 c o s x d x x 2d s in x
x 2 sin x 2 x sin xdx
x 2 sin x 2 xd cos x
x 2 sin x 2 x cos x 2 sin x c .
3. x c . 4. tan t . f (x)
5. x ln( x 1 x 2 ) 1 x 2 c.
6. 2e x ( x 1) c . 7. 1 e2x2 c . 4
8. 1 x 2 f ( x 2 ) 1 f ( x 2 ) c .
2
2
p 5 9 .二 .1 . e sin x s in 2 x d x 2 e sin x s in x d s in x
s in 2 x d x u (1 c o s 2 u )d u
u2 2
1 2
u
d
sin 2 u
u2 1
1
2 2 u sin 2 u 2 sin 2 u du
u2 1
1
u sin 2 u cos 2 u c
22
4
x 1 x sin 2 x 1 cos 2 x c .
22
dx
x2
4 1 x2
1 2
1 d 2 1 arcsin 2 c .
1
4 x2
x
2
x
解 法 (2) 令 x 2 sec t,
x
dx x2
4
1 dt 2
1t 2
c
1 arccos 2 c.
2
x
p54. 3. 令 x 1 t6,
dx
6t5
t3
x1 3 x1
t3 t2dt 6
dt t1
p57.一 .1.
dx x2 2x 3
d ( x 1) ( x 1)2 22
1 ln x 3 c 4 x1
2.
(x2
dx 1)( x 2
3)
1 2
1 ( x2 1
1 x2
)dx 3
1 arctan x 1 arctan x c.
2
23
3
dx
d ( x 1) 1
( x 1)
( s e c 6 x 2 s e c 4 x s e c 2 x )d s e c x
1 sec7 x 2 sec5 x 1 sec3 x c.
7
5
3
p54.sin sin 1 [cos( ) cos( )] 2
cos cos 1 [cos( ) cos( )] 2
32
xdx
11
3. ( x 1)2 ( x 2 1) 2( x 1) 2 arctan x c.
注 意 ( x 2 1) ( x 1)2 2 x .
p 5 8 .三 .1 .
3
dx cos
x
2
dx 2 cos2
x
2
dx
2 co s2 x (1 sec 2 x )
2
2
d tan x
4
p56. 5. 当 x 0时 ,
x e | x |d x x e x d x x e x e x c 1
e x( x 1) c1; 当 x 0时 ,
x e | x |d x x e x d x x e x e x c 2
e x(x 1) c2; 原函数连续可微,
1 x3
2c
x1
3 3
p59.3. e 2 x3 dx u 2 x 3 e uudu ude u
ueu eu ( 2 x 3 1)e 2x3 c.
p60.4. sin(ln x)dx x sin(ln x) cos(ln x)dx
x sin(ln x) x cos(ln x) sin(ln x)dx,
2
2 tan 2 x
2
1
tan x arctan 2 c.
2
2
p58.
2
.
1
sin
dx x
c
o
s
x
dx
2 cos2 x 2 sin x cos x
2
22
d tan x
1
2 tan x
ln | 1 tan
x 2
| c.
2
3 .
1
dx u x 4
x4x
4 u 3du u2 u
4
1 2x2
1 arctan 2
x2
c.
解 法 2:I
1
x
3(
x4
dx 1)
x
x
4
(
x
4
dx 1)
1 x3
dx
x x4
dx 1
1 2x2
1 2
arctan
x2
c.
§3.3不定积分的分部积分法(55-56)
一 .1. 2 x 2e x2 e x2 c
.2.
x2 ln x
x2 c .
(u 1 1 )du u1
p58. 2u 2 4u 4 ln(u 1) c
2 x 4 4 x 4 ln(1 4 x ) c.
4.
1 1
x x
dx
1 x dx
1 x2
1
x
1 x2 dx
dx 1 x2
arcsin x 1 x 2 c.
§习题课(59-60)
一 .1. 2 x(ln x 1) c . 2. cos x 2 sin x c. x
一 .1. e e x c , ln | ln x | c .
2. ln | x sin x | sin x 1 sin5 x 2 sin 3 x c .
5
3
ln | sin x cos x | c
3.
I
(sin
x
cos
x )n2
n2
c
n 2 .
n 2
4. I x 2 x4 c . 5. x 2 sin t, x 3 sec t . 4
5.
I
1 2
(cos 2
x
c o s 1 2 x )d x
1 sin 2 x 1 sin 12 x c.
4
24
6.
( x 3)dx x2 2x 5
1 2
(2 x 2)dx x2 2x 5
x2
2dx 2x
5
1 ln ( x 2 2 x 5) arctan x 1 c.